To make it conceptually clearer, imagine \(\displaystyle \mathbb{K} = \mathbb{C}\) and \(\displaystyle \mathbb{K}_0 = \mathbb{R}\).

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If \(\displaystyle f(x,y) \in \mathbb{R}\[x, y\]\) has degree d which defines an algebraic curve \(\displaystyle \Gamma \subset \mathbb{C}^2\) when considering solutions to \(\displaystyle f(x,y) = 0\) in \(\displaystyle \mathbb{C}^2\),

and \(\displaystyle y - mx - b \in \mathbb{R}\[x, y\]\) is polynomial defining the line \(\displaystyle L \subset \mathbb{C}^2\) when considering solutions to \(\displaystyle y - mx - b = 0\) in \(\displaystyle \mathbb{C}^2\),

then the points of \(\displaystyle (x,y) \in \mathbb{C}^2\) which are on the intersection of \(\displaystyle \Gamma\) and \(\displaystyle L\) will have x-values satisfying \(\displaystyle f(x, mx+b) = 0\) (and then corresponding y-values via \(\displaystyle y = mx - b\)).

Let \(\displaystyle p(x) = f(x, mx+b)\). Then since \(\displaystyle m, b \in \mathbb{R}\) and \(\displaystyle f(x,y) \in \mathbb{R}\[x, y\]\), have that \(\displaystyle p(x) \in \mathbb{R}\[x\]\).

Since \(\displaystyle f\) has degree d, so does p.

If \(\displaystyle \{(x_k, y_k)\}_{k = 0}^d\) are the d intersection points (including possibly multiplicity) of \(\displaystyle \Gamma\) and \(\displaystyle L\), have that \(\displaystyle p(x) = c \prod_{k = 1}^{d}(x - x_k) \in \mathbb{C}\[x\]\).

So since \(\displaystyle p\) has degree d, and is in \(\displaystyle \mathbb{R}\[x\]\), if the first (d-1) of those \(\displaystyle x_k\) are in \(\displaystyle \mathbb{R}\), can you prove that the d-th one, the final one, \(\displaystyle x_d\), is also in \(\displaystyle \mathbb{R}\)?

That's the point of this exercise.

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There's another possibility to consider: \(\displaystyle y - mx - b \in \mathbb{C}\[x, y\]\) might not be capable of describing the line \(\displaystyle L\), in which case you'll need: \(\displaystyle x - s \in \mathbb{C}\[x, y\]\) to describe \(\displaystyle L\).

But since the min poly of \(\displaystyle L\) is in \(\displaystyle \mathbb{R}\[x, y\]\), that tells you that \(\displaystyle s \in \mathbb{R}\). In this case, points of intersection will satisfy \(\displaystyle f(s, y) = 0\),

and so you'll be using a polynomial \(\displaystyle q(y) = f(s, y) \in \mathbb{R}\[y\]\) to do the same reasoning as with above, except now using \(\displaystyle y\)'s and \(\displaystyle q(y)\) instead of \(\displaystyle x\)'s and \(\displaystyle p(x)\).

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The reasoning for general alg closed \(\displaystyle \mathbb{K}\) with subfield \(\displaystyle \mathbb{K}_0\) will work exactly the same way.